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ANTONII [103]
3 years ago
6

An employee at a health club takes home an after-tax salary of $950 every two weeks. If the tax rate is 8%, how much does the em

ployee make every two weeks before taxes? Round your answer to the nearest dollar.Write an equation and explain how you used it to find your answer.
Mathematics
1 answer:
ycow [4]3 years ago
3 0

Answer:

$1032.60

Step-by-step explanation:

Given data

salary = $950 every two weeks

tax rate = 8%

let the initial pay be x

and let the pay after tax deduction be p

p= x-8/100x

Hence the equation for his pay is

p=x-0.08x-------------1

subsitute

950=x-0.08x

950= 0.92x

divide both sides by 0.92

x= 950/0.92

x=$1032.60

The employee make $1032.60 every two weeks before taxes

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Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen
Jobisdone [24]

Parameterize S by the vector function

\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k

so that the normal vector to S is given by

\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k

with magnitude

\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}

In this case, the normal vector is

\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k

with magnitude \sqrt{1^2+2^2+1^2}=\sqrt6. The integral of f(x,y,z)=e^z over S is then

\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx

where T is the region in the x,y plane over which S is defined. In this case, it's the triangle in the plane z=0 which we can capture with 0\le x\le8 and 0\le y\le\frac{8-x}2, so that we have

\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}

5 0
3 years ago
Scores on a test are normally distributed with a mean of 81.2 and a standard deviation of 3.6. What is the probability of a rand
Misha Larkins [42]

<u>Answer:</u>

The probability of a randomly selected student scoring in between 77.6 and 88.4 is 0.8185.

<u>Solution:</u>

Given, Scores on a test are normally distributed with a mean of 81.2  

And a standard deviation of 3.6.  

We have to find What is the probability of a randomly selected student scoring between 77.6 and 88.4?

For that we are going to subtract probability of getting more than 88.4 from probability of getting more than 77.6  

Now probability of getting more than 88.4 = 1 - area of z – score of 88.4

\mathrm{Now}, \mathrm{z}-\mathrm{score}=\frac{88.4-\mathrm{mean}}{\text {standard deviation}}=\frac{88.4-81.2}{3.6}=\frac{7.2}{3.6}=2

So, probability of getting more than 88.4 = 1 – area of z- score(2)

= 1 – 0.9772 [using z table values]

= 0.0228.

Now probability of getting more than 77.6 = 1 - area of z – score of 77.6

\mathrm{Now}, \mathrm{z}-\text { score }=\frac{77.6-\text { mean }}{\text { standard deviation }}=\frac{77.6-81.2}{3.6}=\frac{-3.6}{3.6}=-1

So, probability of getting more than 77.6 = 1 – area of z- score(-1)

= 1 – 0.1587 [Using z table values]

= 0.8413

Now, probability of getting in between 77.6 and 88.4 = 0.8413 – 0.0228 = 0.8185

Hence, the probability of a randomly selected student getting in between 77.6 and 88.4 is 0.8185.

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3 years ago
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4*10^5 where 5 is the unknown exponent
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3 years ago
Please help
Zolol [24]

Given:

The function is

r(x)=0.05(x^2+1)(x-6)

where, function r gives the instantaneous growth rate of a fruit fly population x days after the start of an experiment.

To find:

Number of complex and real zeros.

Time intervals for which the population increased and population deceased.

Solution:

We have,

r(x)=0.05(x^2+1)(x-6)

r(x)=0.05(x^3+x-6x^2-6)

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From the given graph it is clear that, the graph of function r(x) intersect x-axis at once.

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Therefore, function r has 2 complex zeros and one real zero.

Before x=6, the graph of r(x) is below the x-axis and after that the graph of r(x) is above the x-axis.

Negative values of r(x) represents the decrease in population and positive value of r(x) represents the increase in population.

Therefore, based on instantaneous growth rate, the population decreased between 0 and 6 hours and the population increased after 6 hours.

3 0
3 years ago
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